data compressionferragin/teach/information... · 2015-12-15 · prof. paolo ferragina, algoritmi...

Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"

Data Compression

Basics + Huffman coding

How much can we compress?

Assuming all input messages are valid, if even one string

is (lossless) compressed, some other must expand.

Take all messages of length n.

Is it possible to compress ALL OF THEM in less bits ?

NO, they are 2n but we have less compressed msg…

2221

1

−=∑−

=

nn

i

i

We need to talkabout stochastic sources


Entropy (Shannon, 1948)

For a set of symbols S with probability p(s), the self information of s is:

bits

Lower probability � higher information

Entropy is the weighted average of i(s)

∑∈

∗=Ss sp

spSH)(

1log)()( 2

)(log)(

1log)( 22 sp

spsi −==

Statistical Coding

How do we use probability p(s) to encode s?

� Prefix codes and relationship to Entropy

� Huffman codes

� Arithmetic codes


Uniquely Decodable Codes

A variable length code assigns a bit string

(codeword) of variable length to every symbol

e.g. a = 1, b = 01, c = 101, d = 011

What if you get the sequence 1011 ?

A uniquely decodable code can always be

uniquely decomposed into their codewords.

Prefix Codes

A prefix code is a variable length code in which

no codeword is a prefix of another one

e.g a = 0, b = 100, c = 101, d = 11

Can be viewed as a binary trie

0 1

a

b c

d

0

0 1

1


Average Length

For a code C with codeword length L[s], the

average length is defined as

We say that a prefix code C is optimal if for

all prefix codes C’, La(C) ≤ La(C’)

∑∈

∗=Ss

a sLspCL ][)()(

A property of optimal codes

Theorem (Kraft-McMillan). For any optimal

uniquely decodable code, it does exist a prefix

code with the same symbol lengths and thus

same average optimal length. And vice versa…

Theorem (golden rule). If C is an optimal prefix

code for a source with probabilities {p1, …, pn}

then pi < pj � L[si] ≥ L[sj]


Relationship to Entropy

Theorem (lower bound, Shannon). For any probability distribution and any uniquely decodable code C, we have

Theorem (upper bound, Shannon). For any probability distribution, there exists a prefix code C such that

)()( CLSH a≤

1)()( +< SHCLa

Shannon code

takes log 1/p bits

Huffman Codes

Invented by Huffman as a class assignment in ‘50.

Used in most compression algorithms

� gzip, bzip, jpeg (as option), fax compression,…

Properties:

� Generates optimal prefix codes

� Cheap to encode and decode

� La(Huff) = H if probabilities are powers of 2

� Otherwise, at most 1 bit more per symbol!!!


Running Example

p(a) = .1, p(b) = .2, p(c ) = .2, p(d) = .5a(.1) b(.2) d(.5)c(.2)

a=000, b=001, c=01, d=1There are 2n-1 “equivalent” Huffman trees

(.3)

(.5)

(1)

What about ties (and thus, tree depth) ?

0

0

0

11

1

Encoding and Decoding

Encoding: Emit the root-to-leaf path leading to the symbol to be encoded.

Decoding: Start at root and take branch for each bit received. When at leaf, output its symbol and return to root.

a(.1) b(.2)

(.3) c(.2)

(.5) d(.5)0

0

0

1

1

1

abc... �00000101

101001... � dcb


A property on tree contraction

...by induction, optimality follows…

Something like substituting symbols x,y with one new symbol x+y

Optimum vs. Huffman


Model size may be large

Huffman codes can be made succinct in the representation of

the codeword tree, and fast in (de)coding.

We store for any level L:

� firstcode[L]

� Symbol[L,i], for each i in level L

This is ≤ h2+ |Σ| log |Σ| bits

Canonical Huffman tree

= 00.....0

Canonical HuffmanEncoding

1 2 3 4 5


Canonical Huffman

Decoding

T=...00010...

firstcode[1]=2firstcode[2]=1firstcode[3]=1firstcode[4]=2firstcode[5]=0

Problem with Huffman Coding

Consider a symbol with probability .999. The self information is

If we were to send 1000 such symbols we might hope to use 1000*.0014 = 1.44 bits.

Using Huffman, we take at least 1 bit per symbol, so we would require 1000 bits.

00144.)999log(. =−


What can we do?

Macro-symbol = block of k symbols

☺ 1 extra bit per macro-symbol = 1/k extra-bits per symbol

� Larger model to be transmitted

Shannon took infinite sequences, and k � ∞ !!

In practice, we have:

� Model takes |Σ|k (k * log |Σ|) + h2 (where h might be |Σ|)

� It is H0(SL) ≤ L * Hk(S) + O(k * log |Σ|), for each k ≤ L

Compress + Search ? [Moura et al, 98]

Compressed text derived from a word-based Huffman:

� Symbols of the huffman tree are the words of T

� The Huffman tree has fan-out 128

� Codewords are byte-aligned and tagged

1 0 0

Byte-aligned codeword

γ α βtagging

“or”7 bits

Codeword

γ α βhuffman

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]

T = “bzip or not bzip”

β1

[ ]

α α

α αC(T)

α

α

α

βγ

γ

βbzip

β space

or not


CGrep and other ideas...

yes no

no

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]

T = “bzip or not bzip”

β1

[ ]

α α

α αC(T)

P= bzip

yes

α

α

α

βγ

γ

βbzip

β space

or not

= 1α 0β

Byte

-search

Speed ≈ Compression ratio

GREP

You find this at

You find it under my Software projects


Data Compression

Basic search algorithms:Single and Multiple-patterns

Mismatches and Edits

Problem 1

yes no

no

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]

S = “bzip or not bzip”

β1

[ ]

α α

α αC(S)

P = bzip

yes

α

α

α

βγ

γ

βbzip

β space

or not

= 1α 0β


bzip

not

or

space

Dictionary


Pattern Matching Problem

Exact match problem: we want to find all the occurrences of the pattern P[1,m] in the text T[1,n].

T

P

BDAB C ABA

BA

� Naïve solution� For any position i of T, check if T[i,i+m-1]=P[1,m]� Complexity: O(nm) time

� (Classical) Optimal solutions based on comparisons

� Knuth-Morris-Pratt

� Boyer-Moore

☺ Complexity: O(n + m) time

Semi-numerical pattern matching

� We show methods in which Arithmetic and Bit-operations replace comparisons

� We will survey two examples of such methods

� The Random Fingerprint method due to Karp and Rabin

� The Shift-And method due to Baeza-Yates and Gonnet


Rabin-Karp Fingerprint

� We will use a class of functions from strings to integers in order to obtain:

� An efficient randomized algorithm that makes an error with

small probability.

� A randomized algorithm that never errors whose running time is efficient with high probability.

� We will consider a binary alphabet. (i.e., T={0,1}n)

Arithmetic replaces Comparisons

� Strings are also numbers, H: strings → numbers.

� Let s be a string of length m

� P = 0 1 0 1 H(P) = 230+221+210+201= 5

� s=s’ if and only if H(s)=H(s’ )

� Definition:

let Tr denote the m length substring of T starting at

position r (i.e., Tr = T[r,r+m-1]).

∑ =−= m

i

im issH1

][2)(



� Strings are also numbers, H: strings → numbers� Exact match = Scan T and compare H(Ti) and H(P)

� There is an occurrence of P starting at position r of T if and only if H(P) = H(Tr)

T = 1 0 1 1 0 1 0 1

P = 0 1 0 1 H(P) = 5

T = 1 0 1 1 0 1 0 1 H(T2) = 6 ≠ H(P)

P = 0 1 0 1

T = 1 0 1 1 0 1 0 1 H(T5) = 5 = H(P) Match!

P = 0 1 0 1


� We can compute H(Tr) from H(Tr-1)

T = 1 0 1 1 0 1 0 1

T1 = 1 0 1 1

T2 = 0 1 1 0

)1()1(2)(2)( 1 −++−−= − nrTrTTHTH mrr

)0110(61622012112)0110()(

11)1011()(4

2

1

HHTH

HTH

==−=+⋅−⋅==

==



� A simple efficient algorithm:

� Compute H(P) and H(T1)

� Run over T

� Compute H(Tr) from H(Tr-1) in constant time,and make the comparisons (i.e., H(P)=H(Tr)).

� Total running time O(n+m)?

� NO! why?

� The problem is that when m is large, it is unreasonable to assume that each arithmetic operation can be done in O(1) time.

� Values of H() are m-bits long numbers. In general, they are too BIG to fit in a machine’s word.

� IDEA! Let’s use modular arithmetic:

For some prime q, the Karp-Rabin fingerprint of a string s is defined by Hq(s) = H(s) (mod q)

An example

P = 1 0 1 1 1 1 H(P) = 47

q = 7 Hq(P) = 47 (mod 7) = 5

Hq(P) can be computed incrementally!

Intermediate values are also small! (< 2q)

)(5)7(mod5

51)7(mod22

21)7(mod24

41)7(mod25

51)7(mod22

20)7(mod21

PHq===+⋅=+⋅=+⋅=+⋅=+⋅

We can still compute Hq(Tr) from Hq(Tr-1).

2m (mod q) = 2(2m-1 (mod q)) (mod q)


Karp-Rabin Fingerprint

� How about the comparisons?

Arithmetic:There is an occurrence of P starting at position r of T if and only if H(P) = H(Tr)

Modular arithmetic:If there is an occurrence of P starting at position r of T then Hq(P) = Hq(Tr)

False match! There are values of q for which the converse is

not true (i.e., P ≠ Tr AND Hq(P) = Hq(Tr))!

� Our goal will be to choose a modulus q such that� q is small enough to keep computations efficient. (i.e., Hq()s fit in a

machine word)� q is large enough so that the probability of a false match is kept small

Karp-Rabin fingerprint algorithm

� Choose a positive integer I� Pick a random prime q less than or equal to I, and

compute P’s fingerprint – Hq(P).� For each position r in T, compute Hq(Tr) and test to see if

it equals Hq(P). If the numbers are equal either � declare a probable match (randomized algorithm). � or check and declare a definite match (deterministic

algorithm)

� Running time: excluding verification O(n+m).

� Randomized algorithm is correct w.h.p� Deterministic algorithm whose expected running time is

O(n+m)

Proof on the board


Problem 1: Solution

yes

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]


β1

[ ]

α α

α αC(S)

P = bzip

yes

α

α

α

βγ

γ

βbzip

β space

or not

= 1α 0β


bzip

not

or

space

Dictionary

Karp

-Rabin

The Shift-And method

� Define M to be a binary m by n matrix such that:

M(i,j) = 1 iff the first i characters of P exactly match the icharacters of T ending at character j.

� i.e., M(i,j) = 1 iff P[1 … i] = T[j-i+1...j]

� Example: T = california and P = for

� How does M solve the exact match problem?

M

r

o

f

ainrofilac

000100O0003

00001000002

00000100001

10

987654321n

m *****

*****

j

i

T

P


How to construct M

� We want to exploit the bit-parallelism to compute the j-thcolumn of M from the j-1-th one� Machines can perform bit and arithmetic operations between

two words in constant time.� Examples:

� And(A,B) is bit-wise and between A and B. � BitShift(A) is the value derived by shifting the A’s bits down by

one and setting the first bit to 1.

� Let w be the word size. (e.g., 32 or 64 bits). We’ll assume m=w. NOTICE: any column of M fits in a memory word.

=

0

1

1

0

1

)

1

0

1

1

0

(BitShift

How to construct M

� We want to exploit the bit-parallelism tocompute the j-th column of M from the j-th one

� We define the m-length binary vector U(x) for each character x in the alphabet. U(x) is set to 1for the positions in P where character x appears.

� Example:

P = abaac

=

0

1

1

0

1

)(aU

=

0

0

0

1

0

)(bU

=

1

0

0

0

0

)(cU


How to construct M

� Initialize column 0 of M to all zeros

� For j > 0, j-th column is obtained by

� For i > 1, Entry M(i,j) = 1 iff

(1) The first i-1 characters of P match the i-1characters of T

ending at character j-1 ⇔ M(i-1,j-1) = 1

(2) P[i] = T[j] ⇔ the i-th bit of U(T[j]) = 1

� BitShift moves bit M(i-1,j-1) in i-th position

� AND this with i-th bit of U(T[j]) to estabilish if both are true

])[(&))1(()( jTUjMBitShiftjM −=

An example j=1

1 2 3 4 5 6 7 8 9 10

T = x a b x a b a a c a

1 2 3 4 5

P = a b a a c

03

04

5

2

1

n

m

0

0

0

10

987654321

=

0

0

0

0

0

)(xU

=

=

0

0

0

0

0

0

0

0

0

0

&

0

0

0

0

1

])1[(&))0(( TUMBitShift


An example j=2

1 2 3 4 5 6 7 8 9 10


1 2 3 4 5

P = a b a a c

=

0

1

1

0

1

)(aU

003

004

5

2

1

n

m

00

00

10

10

987654321

=

=

0

0

0

0

1

0

1

1

0

1

&

0

0

0

0

1


An example j=3

1 2 3 4 5 6 7 8 9 10


1 2 3 4 5

P = a b a a c

=

0

0

0

1

0

)(bU

0003

0004

5

2

1

n

m

100

000

010

10

987654321

=

=

0

0

0

1

0

0

0

0

1

0

&

0

0

0

1

1



An example j=9

1 2 3 4 5 6 7 8 9 10


1 2 3 4 5

P = a b a a c

=

1

0

0

0

0

)(cU

0010000003

0100000004

5

2

1

n

m

000100100

100000000

011010010

10

987654321

=

=

1

0

0

0

0

1

0

0

0

0

&

1

0

0

1

1


Shift-And method: Complexity

� If m<=w, any column and vector U() fit in a memory word.

� Any step requires O(1) time.

� If m>w, any column and vector U() can bedivided in m/w memory words.

� Any step requires O(m/w) time.

� Overall O(n(1+m/w)+m) time.

� Thus, it is very fast when pattern length is closeto the word size.

� Very often in practice. Recall that w=64 bits in modern architectures.


Some simple extensions

� We want to allow the pattern to contain special symbols, like [a-f] classes of chars

=

0

1

1

0

1

)(aU

=

0

0

0

1

1

)(bU

=

1

0

0

0

0

)(cU

P = [a-b]baac

� What about ‘?’, ‘[^…]’ (not).

Problem 1: An other solution

yes no

no

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]


β1

[ ]

α α

α αC(S)

P = bzip

yes

α

α

α

βγ

γ

βbzip

β space

or not

= 1α 0β


bzip

not

or

space

Dictionary

Shift-A

nd


Problem 2

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]


β1

[ ]

α α

α αC(S)

P = o

yes

α

α

α

βγ

γ

βbzip

β space

or not

Speed ≈ Compression ratio?

bzip

not

or

space

DictionaryGiven a pattern P findall the occurrences in Sof all terms containingP as substring Codes

not= 1 γ 0 γ 0 α

or = 1 γ 0 α 0 βyes

No! Why?

Shift-And

Shift-And

A scan of C(s) for each term that contains P

Multi-Pattern Matching Problem

Given a set of patterns PPPP = {P1,P2,…,Pl} of total length m, we want

to find all the occurrences of those patterns in the text T[1,n].

T

P1

BDAB C ABA

AC

� Naïve solution� Use an (optimal) Exact Matching Algorithm searching eachpattern in PPPP

� Complexity: O(nl+m) time, not good with many patterns

� Optimal solution due to Aho and Corasick

� Complexity: O(n + l + m) time

P2 AD


A simple extention of Shift-And

� S is the concatenation of the patterns in P

� R is a bitmap of lenght m.� R[i] = 1 iff S[i] is the first symbol of a pattern

� Use a variant of Shift-And method searching forS� For any symbol c, U’(c) = U(c) and R

� U’(c)[i] = 1iff S[i]=c and is the first symbol of a pattern

� For any step j,

� compute M(j)

� then M(j) OR U’(T[j]). Why?

� Set to 1 the first bit of each pattern that start withT[j]

� Check if there are occurrences ending in j. How?

Problem 3

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]


β1

[ ]

α α

α αC(S)

P = bot k=2

α

α

α

βγ

γ

βbzip

β space

or not

bzip

not

or

space

DictionaryGiven a pattern P findall the occurrences in Sof all terms containingP as substring allowingat most k mismatches

???


Agrep: Shift-And method with errors

� We extend the Shift-And method for finding inexact occurrences of a pattern in a text.

� Example:T = aatatccacaaP = atcgaa

P appears in T with 2 mismatches starting at position 4, it also occurs with 4 mismatches starting at position 2.

a a t a t c c a c a a a a t a t c c a c a a

a t c g a a a t c g a a

Agrep

� Our current goal given k find all the occurrences of P in T with up to k mismatches

� We define the matrix Ml to be an m by n binary matrix, such that:

Ml(i,j) = 1 iffThere are no more than l mismatches between the first i characters of P match the i characters up through character j of T.

� What is M0?

� How does Mk solve the k-mismatch problem?


Computing Mk

� We compute Ml for all l=0, … ,k.

� For each j compute M(j), M1(j), … , Mk(j)

� For all l initialize Ml(0) to the zero vector.

� In order to compute Ml(j), we observe that there is a match iff

Computing Ml: case 1

� The first i-1 characters of P match a substring of T ending at j-1, with at most l mismatches, and the next pair of characters in P and T are equal.

*****

*****i-1

])[())1(( jTUjMBitShift l ∧−


Computing Ml: case 2

� The first i-1 characters of P match a substring of T ending at j-1, with at most l-1 mismatches.

*****

*****

j-1

i-1

))1(( 1 −− jMBitShift l

Computing Ml

� We compute Ml for all l=0, … ,k.

� For each j compute M(j), M1(j), … , Mk(j)

� For all l initialize Ml(0) to the zero vector.

� In order to compute Ml(j), we observe that there is a match iff

))1((

))](())1(([

)(

1 −∨∧−

=

− jMBitShift

jTUjMBitShift

jM

l

l

l


Example M1

1 2 3 4 5 6 7 8 910


P = a b a a d

10010010003

00100100004

5

2

1

0110100100

0100000000

1111111111

10

987654321

00010000003

00100000004

5

2

1

0000100100

0000000000

1011010010

10

987654321M0=

M1=

How much do we pay?

� The running time is O(kn(1+m/w)

� Again, the method is practically efficient for small m.

� Still only a O(k) columns of M are needed at any given time. Hence, the space used by the algorithm is O(k) memory words.


Problem 3: Solution

β0

γ γ0 0

β1 1

1

γ 0 β01

β1 β01

[bzip] [ ]

[bzip][ ] [not]

[or]


β1

[ ]

α α

α αC(S)

P = bot k=2

yes

α

α

α

βγ

γ

βbzip

β space

or not

bzip

not

or

space

DictionaryGiven a pattern P findall the occurrences in Sof all terms containingP as substring allowingk mismatches

Codes

not= 1 γ 0 γ 0 αyes

Agrep

Shift-And

Agrep: more sophisticated operations

� The Shift-And method can solve other ops

� The edit distance between two strings p and s is

d(p,s) = minimum numbers of operations

needed to transform p into s via three ops:

� Insertion: insert a symbol in p

� Delection: delete a symbol from p

� Substitution: change a symbol in p with a different one

� Example: d(ananas,banane) = 3

� Search by regular expressions

� Example: (a|b)?(abc|a)


Data Compression

Some thoughts

Variations…

Canonical Huffman still needs to know the codeword lengths, and thus build the tree…

This may be extremely time/space costly when you deal with Gbs of textual data

Sort pi in decreasing order, and encode si via

the variable-length code for the integer i.

A simple algorithm


γ−code for integer encoding

� x > 0 and Length = log2 x +1

e.g., 9 represented as <000,1001>.

� γ−code for x takes 2 log2 x +1 bits (ie. factor of 2 from optimal)

0000...........0 x in binary Length-1

� Optimal for Pr(x) = 1/2x2, and i.i.d integers

It is a prefix-free encoding…

� Given the following sequence of γ−coded integers, reconstruct the original sequence:

0001000001100110000011101100111

8 6 3 59 7


Analysis

Sort pi in decreasing order, and encode si via

the variable-length code γ(i).

Recall that: |γ(i)| ≤ 2 * log i + 1

How much good is this approach wrt Huffman?

Compression ratio ≤ 2 * H0(s) + 1

Key fact:

1 ≥Σi=1,...,x pi ≥ x * px � x ≤ 1/px

How good is it ?

Encode the integers via δ-coding:

|γ(i)| ≤ 2 * log i + 1

The cost of the encoding is (recall i ≤ 1/pi):

∑∑Σ=Σ=

+∗≤∗,..,1,...,1

]11

log*2[|)(|i i

ii

i ppip γ

This is: 1)(*2 0 +≤ XH

No much worse than Huffman,

and improvable to H0(X) + 2 + ....


A better encoding

� Byte-aligned and tagged Huffman

� 128-ary Huffman tree

� First bit of the first byte is tagged

� Configurations on 7-bits: just those of Huffman

� End-tagged dense code

� The rank r is mapped to r-th binary sequence on 7*k bits

� First bit of the last byte is tagged

A better encoding

Surprising changes� It is a prefix-code

� Better compression: it uses all 7-bits configurations


(s,c)-dense codes

Distribution of words is skewed: 1/iθ, where 1 < θ < 2

� A new concept: Continuers vs Stoppers

� Previously we used: s = c = 128

� The main idea is:� s + c = 256 (we are playing with 8 bits)� Thus s items are encoded with 1 byte� And s*c with 2 bytes, s * c2 on 3 bytes, ...

� An example� 5000 distinct words

� ETDC encodes 128 + 1282 = 16512 words on 2 bytes

� (230,26)-dense code encodes 230 + 230*26 = 6210 on 2 bytes, hence more on 1 byte and thus if skewed...

It is

a pre

fix-co

de

Optimal (s,c)-dense codes

Find the optimal s, by assuming c = 128-s.

� Brute-force approach

� Binary search:� On real distributions, it seems that one unique minimum

Ks = max codeword length

Fsk = cum. prob. symb. whose |cw| <= k


Experiments: (s,c)-DC much interesting…

Search is 6% faster than byte-aligned Huffword

Streaming compression

Still you need to determine and sort all terms….

Can we do everything in one pass ?

� Move-to-Front (MTF):

� As a freq-sorting approximator

� As a caching strategy

� As a compressor

� Run-Length-Encoding (RLE):

� FAX compression


Move to Front Coding

Transforms a char sequence into an integersequence, that can then be var-length coded

� Start with the list of symbols L=[a,b,c,d,…]

� For each input symbol s

1) output the position of s in L

2) move s to the front of L

Properties:

� Exploit temporal locality, and it is dynamic

� X = 1n 2n 3n… nn � Huff = O(n2 log n), MTF = O(n log n) + n2

There is a memory

MTF: how good is it ?

Encode the integers via δ-coding:

|γ(i)| ≤ 2 * log i + 1

Put Σ in the front and consider the cost of encoding:

∑∑Σ

= =−

−+ΣΣ1 2

)()log(1

x

n

i

xxx

iippO γ

∑Σ

=

++ΣΣ≤1

]1log*2[)log(x x

x n

NnOBy Jensen’s:

]1)(*2[*)log( 0 ++ΣΣ≤ XHNO

No much worse than Huffman

...but it may be far better

)1()(*2][ 0 OXHmtfLa +≤


MTF: higher compression

Alphabet of words

How to keep efficiently the MTF-list:

� Search tree

� Leaves contain the words, ordered as in the MTF-List

� Nodes contain the size of their descending subtree

� Hash Table

� keys are the words (of the MTF-List)

� data is a pointer to the corresponding tree leaves

� Each ops takes O(log Σ)

� Total cost is O(n log Σ)

Run Length Encoding (RLE)

If spatial locality is very high, then

abbbaacccca => (a,1),(b,3),(a,2),(c,4),(a,1)

In case of binary strings � just numbers and one bit

Properties:

� Exploit spatial locality, and it is a dynamic code

� X = 1n 2n 3n… nn �

Huff(X) = n2 log n > Rle(X) = n (1+log n)

There is a memory

data compressionferragin/teach/information... · 2015-12-15 · prof. paolo ferragina, algoritmi...

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